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intigret cos2x.sin4x/cos^4x(1+cos^2 2x)

intigret cos2x.sin4x/cos^4x(1+cos^2 2x)

Grade:12

2 Answers

Y RAJYALAKSHMI
45 Points
6 years ago
cos2x sin4x/[cos4x(1+cos2 2x)] can be written as tanx  – 16cos3x sin3x / [cos4x(3+cos4x)]
 
∫tanx dx = sec2x
put cos4x/(3+cos4x) = t
then – 16cos3xsin3x/(3+cos4x)2 dx = dt
∫– 16cos3x sin3x / [cos4x(3+cos4x)]dx = ∫1/t dt = log t = log [cos4x/(3+cos4x)] = 4log(cosx) – log(3 + cos4x)
 
So, the integral of given function is sec2x + 4log(cosx) – log(3 + cos4x) + C
 
 
 
 
 
Y RAJYALAKSHMI
45 Points
6 years ago
Sorry,
 
Pl. ignore, this is a wrong solution

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